the side QR of ∆PQR is produced a point S .if the bisector of ∆PQR and ∆PRS meet at point T , then prove that ∆QTR=1\2∆QPR.
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bhoomi81:
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Hello mate ☺
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Solution:
∠PQT=∠TQR (Given)
∠PRT=∠TRS (Given)
To Prove: ∠QTR=1/2(∠QPR)
∠PRS=∠QPR+∠PQR
(If a side of a triangle is produced, then the exterior angle is equal to the sum of two interior opposite angles.)
⇒∠QPR=∠PRS−∠PQR
⇒∠QPR=2∠TRS−2∠TQR
⇒∠QPR=2(∠TRS−∠TQR)
=2(∠TQR+∠QTR−∠TQR) (∠TRS=∠TQR+∠QTR)
(If a side of a triangle is produced, then the exterior angle is equal to the sum of two interior opposite angles.)
⇒∠QPR=2(∠QTR)
⇒∠QTR=1/2(∠QPR)
Hence Proved
I hope, this will help you.☺
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